funcode

Graph

Question to ask:

• is Directed ?

• is Connected ? or multiple component

• is Cyclic ? or Acyclic ? How many cycle

• is Weighted ? -ve weight allowed ?

• Dense Graph ? Matrix vs AdjList rep

  static class Graph { int V; LinkedList[] adjList;

    Graph(int ver)
    {
        V = ver;
        adjList = new LinkedList[V];
        for (int i = 0; i < V; i++)
            adjList[i] = new LinkedList<>();
    }
  

  }

  static void addEdge(Graph g, int u, int v) { g.adjList[u].add(v); // Add v to u's list. } main() { //Graph g = new Graph(4); //addEdge(g, 0, 1); }

Topological Sort:

• O(V+E): Modify DFS to have one more stack and once DFS of all child are completed, push current node to stack. Stack will have Topological order

• O(V+E): Kahn's Algo: InOrder => Queue with O-in-degree = > Start with 0 and -1 all adjacent and continue

Shortest Path

(source vertex is always present for SP problem)

• unweighted : O(V + E) time modified BFS

• weighted :: Dijkstra’s algorithm (Greedy) O((V+E)LogV) (Use min-PQ for edges weight) same for Prim's MST

• Bellman-Ford (Graph with -ve value) O(VE) - Run |V|-1 times loop to update Shortest distance

  • The Vth iteration with -ve value will confirm there is a cycle.

• For DAG: O(V+E) simply topological sort and then process in same order and check each adjacent node

  • • DAG: Get Topological Sorting and dist[v] = Min(dist[v], weight(u,v)+dist[u])

• O(V3) Floyd Warshall Algorithm "All Pairs Shortest Path" Three loops of "V"

Longest Path

(w/o source)

• Do DFS from each node, (N * N) and take max. Do DFS from each node with a DP[] and visited[] => O(N)

(w/ a source and weighted, find from source to all other vertices)

• DAG: -1 * Edge_Weight then use Topological Sorting to find shortest path and revert edge value

• DAG: Get Topological Sorting and dist[v] = Max(dist[v], weight(u,v)+dist[u])

(w/ a source and non-weighted)

• Simpy DFS from that source

• If it's a tree, the height path will be longest from root.

Cycle in Graph

• Directed: Do Topological sort and then check if each U,V (parent,child) have order parent < child else cycle(true)

• DFS for a connected graph produces a tree. IF DFS stack already have current node its a cycle.

• DFS using Graph coloring:: WHITE (not processed yet), GREY (Ongoing), BLACK (VISITED)

• UnDirected: DFS and if current node is visited and is not the parent then there is a cycle.

Tree

• Rooted vs Non-Rooted Tree

• LCA: Least(Lowest) common ancestor: Find Path from Root to both A, B and find the junction.

• LCA: Do DFS from root, if root is either A/B return node, if both left/right is non-null return node, else go in one direction.

• BST/AVL/Red-Black: BST can be skewed and worst case can be N, which AVL/RB are BST which enforce balancing

  • AVL is more strict balanced so come with cost of rotation

  • Between tree-map and heap, some time tree-map can be more useful if we need both min/max from with minimal memory footprint.

  • A heap is easily implemented in contiguous memory, i.e. an array. A red-black tree is typically constructed with a separate heap allocation for each node. The red-black tree ends up accessing memory all over the heap for each tree traversal.

  • Heap is simpler and take less memory overhead to get build, while RB BST Tree need pointer and coloring bits per node.

  • TreeMap is based on Red-Black BST Tree: Easy to get Min/Max both using getFirst/getLast key O(1) else O(logN) for all other

  • PQueue- min/max Heap can just tell min/max in O(1) but finding any random number is O(N)

DP

• Place Tiles: Tiles [1,2,4,1] ==> Frequency Map => [1:2, 2:1, 4:1] ==> O(2``N) Brute force f(n) = f(n-1) + f(n-2) for Tiles [1:[]] [2: [][]] For fn(6) => find fn(1), fn(2), fn(3), fn(5), fn(8) DP: For (i:1 => n) => For each tiles length T*m(times) dp[i] += dp[i-T]*m dp[n] ==> Total number of ways to place T tiles over N places.

String

• Longest Increasing Sub-sequence => [1,3,6,2,8,7] ==> 1,3,6,7 (4) Build DAG(N2) and find longest path or use LogN for Binary Search keeping a sub array from left to right

  • Use dp[i] (all "1") If num[i] > num[j] && dp[j] + 1 > dp[i] then dp[i] = dp[j] + 1

  • LIS[i] = MAX(1, 1+ LIS[j]) j < i && num[i] > num[j]

  O(N*N)
  Arrays.fill(dp, 1);
        for (int i = 1; i < nums.length; i++) {
            for (int j = 0; j < i; j++) {
                if (nums[i] > nums[j]) {
                    dp[i] = Math.max(dp[i], dp[j] + 1);
                }
            }
        }
  

• O(N*N) Longest Pallindromic SubString ==> expand around corner (i & i+1)

• Longest Common Seq (LCS) between s1/s2=> N * N matrix and traverse bottom up.

  If Match dp[i][j] = 1 + dp[i-1][j-1]; else max(dp[i-1][j], dp[i][j-1]);

• Longest Pallindromic SubSeq => Reverse "S" and find LCS

• Longest Common SubString between S1/S2 (consecutive)

  If Match dp[i][j] = 1 + dp[i-1][j-1]; else dp[i][j] = 0;